## Position in the Orbit

Although we already know everything about the geometry of the orbit, we still cannot find the planet at a given time, since we do not know the radius vector r as a function of time. The variable in the equation of the orbit is an angle, the true anomaly f, measured from the perihelion. From Kepler's second law it follows that f cannot increase at a constant rate with time. Therefore we need some preparations before we can find the radius vector at a given instant.

The radius vector can be expressed as r = a(cos E - e)i + b sin Ej, (6.34)

wherei and j are unit vectors parallel with the major and minor axes, respectively. The angle E is the eccentric anomaly; its slightly eccentric definition is shown in Fig. 6.9. Many formulas of elliptical motion become very simple if either time or true anomaly is replaced

by the eccentric anomaly. As an example, we take the square of (6.34) to find the distance from the Sun:

= a2 (cos E - e)2 + b2 sin2 E = a2[(cos E - e)2 + (1 - e2)(1 - cos2 E)]

Our next problem is to find how to calculate E for a given moment of time. According to Kepler's second law, the surface velocity is constant. Thus the area of the shaded sector in Fig. 6.10 is

where t — t is the time elapsed since the perihelion, and P is the orbital period. But the area of a part of an ellipse is obtained by reducing the area of the corresponding part of the circumscribed circle by the axial ratio b/a. (As the mathematicians say, an ellipse is an

Fig. 6.10. The area of the shaded sector equals b/a times the area SP'X. S = the Sun, P = the planet, X = the perihelion a sin £

Fig. 6.10. The area of the shaded sector equals b/a times the area SP'X. S = the Sun, P = the planet, X = the perihelion affine transformation of a circle.) Hence the area of SPX is b

By equating these two expressions for the area A, we get the famous Kepler's equation,

is the mean anomaly of the planet at time t. The mean anomaly increases at a constant rate with time. It indicates where the planet would be if it moved in a circular orbit of radius a. For circular orbits all three anomalies f, E, and M are always equal.

If we know the period and the time elapsed after the perihelion, we can use (6.38) to find the mean anomaly. Next we must solve for the eccentric anomaly from Kepler's equation (6.37). Finally the radius vector is given by (6.35). Since the components of r expressed in terms of the true anomaly are r cos f and r sin f, we find cos f = sin f =

These determine the true anomaly, should it be of interest.

Now we know the position in the orbital plane. This must usually be transformed to some other previously selected reference frame. For example, we may want to know the ecliptic longitude and latitude, which can later be used to find the right ascension and declination. These transformations belong to the realm of spherical astronomy and are briefly discussed in Examples 6.5-6.7.

## Telescopes Mastery

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